Question

In: Chemistry

Determine the pH at equivalence when 12.0ml of 0.025mol/L benzoic acid (Ka=6.3x10-5) is titrated with 0.050mol/L...

Determine the pH at equivalence when 12.0ml of 0.025mol/L benzoic acid (Ka=6.3x10-5) is titrated with 0.050mol/L NaOH.

Solutions

Expert Solution

find the volume of NaOH used to reach equivalence point

M(C6H5COOH)*V(C6H5COOH) =M(NaOH)*V(NaOH)

0.025 M *12.0 mL = 0.05M *V(NaOH)

V(NaOH) = 6 mL

we have:

Molarity of C6H5COOH = 0.025 M

Volume of C6H5COOH = 12 mL

Molarity of NaOH = 0.05 M

Volume of NaOH = 6 mL

mol of C6H5COOH = Molarity of C6H5COOH * Volume of C6H5COOH

mol of C6H5COOH = 0.025 M * 12 mL = 0.3 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.05 M * 6 mL = 0.3 mmol

We have:

mol of C6H5COOH = 0.3 mmol

mol of NaOH = 0.3 mmol

0.3 mmol of both will react to form C6H5COO- and H2O

C6H5COO- here is strong base

C6H5COO- formed = 0.3 mmol

Volume of Solution = 12 + 6 = 18 mL

Kb of C6H5COO- = Kw/Ka = 1*10^-14/6.3*10^-5 = 1.587*10^-10

concentration ofC6H5COO-,c = 0.3 mmol/18 mL = 0.0167M

C6H5COO- dissociates as

C6H5COO- + H2O -----> C6H5COOH + OH-

0.0167 0 0

0.0167-x x x

Kb = [C6H5COOH][OH-]/[C6H5COO-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.587*10^-10)*1.667*10^-2) = 1.627*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.627*10^-6 M

[OH-] = x = 1.627*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.627*10^-6)

= 5.79

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.79

= 8.21

Answer: 8.21


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